Question

2. Combined loading (15pts) a) determine the resultants at cross-section a-a (facing positive x-axis). (you may use M-Rx F to determine the resultant moments.) b) Determine stress components at A. 300 00 mm 30 mm 20 mm 200 mmSection a-a 450 N 300 20ON

Answer 1

to the Cross-sehon (Cs) at c 200 N CCw2) Mu 200 N 450N M2

Effect of forces on 'A'

Section properties for the shaft are as follows:
   OD   = 30.0 mm
   ID   = 20.0 mm
   t   = (30.0 mm - 20.0 mm) / 2
       = 5.0 mm
   c   = 30.0 mm / 2
       = 15.0 mm
   Area   = p[(30.0 mm)2 - (20.0 mm)2] / 4
       = 392.7 mm2
   J   = p[(30.0 mm)4 - (20.0 mm)4] / 32
       = 63,813.6 mm4
   I   = p[(30.0 mm)4 - (20.0 mm)4] / 64
       = 31,906.8 mm4
   S   = 31,906.8 mm4 / 15.0 mm
       = 2,127.1 mm3
   Q   = [(30.0 mm)3 - (20.0 mm)3] / 12
       = 1,583.3 mm3

For stress element A (on the top of the shaft):

The force Px= 450.0 N creates the following stresses:

a)   A uniformly distributed axial tension normal stress. The magnitude of the normal stress is given by:

|sx|   = |Nx| / Area
   = 450.0 N / 392.7 mm2
   = 1.146 MPa

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The force Py = -200.0 N creates the following stresses:

a)   A linearly distributed tension normal stress due to a bending moment about the z axis. The magnitude of the normal stress is given by:

|sx|   = |Mz y|/ I
   = (200.0 N)(300.0 mm)(15.0 mm) / 31,906.8 mm4
   = 28.207 MPa

The direction of the stress is shown on the stress element.

b)   Although Py creates shear stress in the shaft, the transverse shear stress on element A in the y direction is zero at this location. When subjected to a shear force in the y direction, the outermost surfaces of the shaft in the y direction are free of shear stress.

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The force Pz = 300.0 N creates the following stresses:

a)   Although Pz creates a moment about the y axis, it produces zero flexural stress on stress element A because z = 0 at this location. In other words, element A is on the neutral axis for moments about the y axis.

b)   A transverse shear stress due to the 300.0 N shear force. The magnitude of the shear stress is given by:

|tV|   = |Vz|Q / I t
   = (300.0 N)(1,583.3 mm3) / [(31,906.8 mm4) (2×5.0 mm)]
   = 1.489 MPa

Sign convention: Generally, the proper shear stress direction is determined by inspection. The direction of the stress is shown on the stress element.

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Summary for stress element A (on the top of the shaft): The normal stresses for the combined loading can be determined by superimposing the individual cases. For stress element A (on the top of the shaft), the total normal stress acting on the element is a tension stress of 29.353 MPa. The shear stresses for the combined loading act in the positive z direction on the positive x face of the element. The magnitude of the shear stress is 1.489 MPa.

The principal stresses for the element are
   sp1 = 29.428 MPa
and
   sp2 = -0.07531 MPa

The maximum in-plane shear stress is
   tmax = 14.752 MPa
and the absolute maximum shear stress equals the in-plane shear stress. This condition occurs when sp1 and sp2 have opposite signs.

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For stress element D (on the +z side of the shaft):

The force Px= 450.0 N creates the following stresses:

a)   A uniformly distributed axial tension normal stress. The magnitude of the normal stress is given by:

|sx|   = |Nx| / Area
   = 450.0 N / 392.7 mm2
   = 1.146 MPa

__________________________________________________

The force Py = -200.0 N creates the following stresses:

a)   Although Py creates a moment about the z axis, it produces zero flexural stress on stress element D because y = 0 at this location. In other words, element D is on the neutral axis for moments about the z axis.

b)   A transverse shear stress due to the 200.0 N shear force. The magnitude of the shear stress is given by:

|tV|   = |Vy|Q / I t
   = (200.0 N)(1,583.3 mm3) / [(31,906.8 mm4) (2×5.0 mm)]
   = 0.9925 MPa

Sign convention: Generally, the proper shear stress direction is determined by inspection. The direction of the stress is shown on the stress element.

__________________________________________________

The force Pz = 300.0 N creates the following stresses:

a)   A linearly distributed compression normal stress due to a bending moment about the y axis. The magnitude of the normal stress is given by:

|sx|   = |My z| / I
   = (300.0 N)(300.0 mm)(15.0 mm) / 31,906.8 mm4
   = 42.311 MPa

The direction of the stress is shown on the stress element.

b)   Although Pz creates shear stress in the shaft, the transverse shear stress on element D in the z direction is zero at this location. When subjected to a shear force in the z direction, the outermost surfaces of the shaft in the z direction are free of shear stress.

__________________________________________________

Summary for stress element D (on the +z side of the shaft): The normal stresses for the combined loading can be determined by superimposing the individual cases. For stress element D (on the +z side of the shaft), the total normal stress acting on the element is a compression stress of 41.165 MPa. The shear stresses for the combined loading act in the negative y direction on the positive x face of the element. The magnitude of the shear stress is 0.9925 MPa.

The principal stresses for the element are
   sp1 = 0.02391 MPa
and
   sp2 = -41.189 MPa

The maximum in-plane shear stress is
   tmax = 20.606 MPa
and the absolute maximum shear stress equals the in-plane shear stress. This condition occurs when sp1 and sp2 have opposite signs.

Effect of Bending on 'A'

My Ơb-Bending stress y- Vertical distance away from the neutral axis I- Moment of inertia around the neutral axis

We know A Bend fos a hollow shup 1 M do do a 30 m0.03 m 32メ40x0.07 36. 4 2-04x10 M Pa 102 42-31